Comments on: Trigonometry – My Grandmother Also Can Do This! http://www.exampaper.com.sg/questions/e-maths/trigonometry-my-grandmother-also-can-do-this Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/trigonometry-my-grandmother-also-can-do-this#comment-112 Miss Loi Mon, 07 May 2007 04:55:54 +0000 http://www.exampaper.com.sg/questions/e-maths/trigonometry-my-grandmother-also-can-do-this#comment-112 Great stuff Nick! The purpose of this question is to remind students that the standard trigo ratios (i.e. TOA CAH SOH -> Tan: Opp/Adj; Cos: Adj/Hypo; Sin: Opp/Hypo) applies only to an <em>acute</em> angle. In order to solve for angle DCB which is an <em>obtuse</em> angle in the question, students need to remember a further set of equations i.e: sin θ = sin (180<sup>o</sup> - θ) cos θ = -cos (180<sup>o</sup> - θ) tan θ = -tan (180<sup>o</sup> - θ) Hence exam questions usually require multiple steps like these. If everyday just simply TOA CAH SOH TOA CAH SOH, the examination board can close shop <i>liao</i>! So remember this well! Great stuff Nick!

The purpose of this question is to remind students that the standard trigo ratios (i.e. TOA CAH SOH -> Tan: Opp/Adj; Cos: Adj/Hypo; Sin: Opp/Hypo) applies only to an acute angle.

In order to solve for angle DCB which is an obtuse angle in the question, students need to remember a further set of equations i.e:

sin θ = sin (180o - θ)
cos θ = -cos (180o - θ)
tan θ = -tan (180o - θ)

Hence exam questions usually require multiple steps like these. If everyday just simply TOA CAH SOH TOA CAH SOH, the examination board can close shop liao!

So remember this well!

]]> By: Nick http://www.exampaper.com.sg/questions/e-maths/trigonometry-my-grandmother-also-can-do-this#comment-108 Nick Sun, 06 May 2007 10:25:29 +0000 http://www.exampaper.com.sg/questions/e-maths/trigonometry-my-grandmother-also-can-do-this#comment-108 sin x = 5/13 => sin ABC = 5/13 => cos BCA = 5/13 => cos DCB = -5/13. sin x = 5/13 => sin ABC = 5/13 => cos BCA = 5/13 => cos DCB = -5/13.

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